perplexus.info :: Probability : Roulette

Roulette (Posted on 2005-03-23)

If I spin the roulette (numbers from 0 to 36) N consecutive times, how many different numbers should I expect to get?

(If N=1, your formula should give 1 as an answer, but for N=2, it should be a little under 2, and for all other N, the formula should be less than the minimum of N and 37.)

Another question: how many times should you spin the roulette, so chances are better than one in a million, that all 37 numbers will have appeared?

Submitted by e.g.

Rating: 4.5000 (2 votes)

Solution: (Hide)

Must check but... D(n)=37(1-(36/37)ⁿ) = (37ⁿ-36ⁿ)/37^n-1. As a check, D(1)=1 and D(2)=73/37... almost 2.
For the second question, the odds that an specific number hasn't appeared after n spins are (36/37)ⁿ, so the odds that any number hasn't appeared after n spins are 37x(36/37)ⁿ ≤10^-6, so n≥636.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
One step needed to get to final solution	Rupesh Khandelwal	2005-04-16 10:22:59
My take on this problem (different solution?, spoilers)	ajosin	2005-03-24 23:15:32
Part 2 spoiler	Charlie	2005-03-23 16:05:43
Hadn't noticed this while it was in the queue, but ... (part 1 spoiler)	Charlie	2005-03-23 15:05:14

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